Electrons fill the electron orbitals (s, p, d, or f) starting from the lowest energy level going to the highest energy level. I hope this helps. let me know if you need more information such as what the

7 0

3 years ago

State whether it would be worthwhile to investigate finding a catalyst to use in this reaction under standard conditions and exp

mash [69]

Yes, it will be worthwhile to investigate finding a catalyst to use in this reaction under standard conditions because it is negative.

<h3>What is a Catalyst?</h3>

This is a substance which speeds up the rate of a chemical reaction by lowering the activation energy.

ΔG being negative indicates a a slow reaction which is why a catalyst under standard conditions should be used.

Read more about Catalyst here brainly.com/question/12507566

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3 0

2 years ago

Glucose, C 6 H 12 O 6 , is used as an energy source by the human body. The overall reaction in the body is described by the equa

MA_775_DIABLO [31]

Answer:

$m_{O_2}=61.87gO_2$

$m_{CO_2}=85.07gCO_2$

Explanation:

Hello,

Considering the given reaction's stoichiometry, grams of oxygen result:

$m_{O_2}=58.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molO_2}{1molC_6H_{12}O_6}*\frac{32gO_2}{1molO_2}\\m_{O_2}=61.87gO_2$

Moreover, the mass of produced carbon dioxide turns out:

$m_{CO_2}=58.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molCO_2}{1molC_6H_{12}O_6}*\frac{44gCO_2}{1molCO_2}\\m_{O_2}=85.07gCO_2$

Best regards.

6 0

3 years ago